In a trapezium-shaped field, one of the parallel sides is twice the other.

Let the lengths of the parallel sides be $x \mathrm{~cm}$ and $2 x \mathrm{~cm}$.

Area of trapezium $=\left\{\frac{1}{2} \times(x+2 x) \times 84\right\} \mathrm{m}^{2}$

$=\left(\frac{1}{2} \times 3 x \times 84\right) \mathrm{m}^{2}$

$=(42 \times 3 x) \mathrm{m}^{2}$

$=126 x \mathrm{~m}^{2}$

Area of the trapezium $=9450 \mathrm{~m}^{2}$ (Given)

$\therefore 126 x=9450$

$\Rightarrow x=\frac{9450}{126}$

$\Rightarrow x=75$

Thus, the length of the parallel sides are $75 \mathrm{~m}$ and $150 \mathrm{~m}$, that is, $(2 \times 75) \mathrm{m}$, and the length of the longer side is $150 \mathrm{~m}$.